Answer:
Explanation:
Let Jacky takes t seconds to catch up with his friend.
Distance traveled by him can be calculated as follows
s =ut + 1/2 a t²
u = 0 , a = 2m / s²
s = .5 x 2 t² = t².
Time during which Davey was in motion is t + 2
Distance covered by him
= 4 ( t + 2 )
These two distances must be equal due to their meeting .
4(t +2 )= t²,
t² -4t -8 =0
t = 5.46 s